In triangle $ABC,$ $AB = 20$ and $BC = 15.$  Find the largest possible value of $\tan A.$
Answer: Consider $A$ and $B$ as fixed points in the plane.  Then the set of possible locations of point $C$ is the circle centered at $B$ with radius 15.

[asy]
unitsize(0.2 cm);

pair A, B, C;

B = (0,0);
A = (20,0);
C = intersectionpoint(arc(B,15,0,180),arc(A,5*sqrt(7),0,180));

draw(A--B--C--cycle);
draw(Circle(B,15), dashed);

label("$A$", A, S);
dot("$B$", B, S);
label("$C$", C, NE);
label("$20$", (A + B)/2, S);
label("$15$", (B + C)/2, NW);
[/asy]

Then $\angle A$ is maximized when $\overline{AC}$ is tangent to the circle.  In this case, $\angle C = 90^\circ,$ so by Pythagoras,
\[AC = \sqrt{20^2 - 15^2} = 5 \sqrt{7}.\]Then $\tan A = \frac{15}{5 \sqrt{7}} = \boxed{\frac{3 \sqrt{7}}{7}}.$